Example 10 - Find term independent of x (3/2x2 - 1/3x)6 - Coefficient

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise
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Example 10 Find the term independent of x in the expansion of ﷐﷐﷐3﷮2﷯﷐𝑥﷮2﷯ − ﷐1﷮3𝑥﷯﷯﷮6﷯,x > 0. Step1 Calculate general term of expansion﷐﷐﷐3﷮2﷯﷐𝑥﷮2﷯ − ﷐1﷮3𝑥﷯﷯﷮6﷯ We know that general term of expansion (a + b)n is Tr+1 = nCr (a)n–r.(a)n For general term of expansion ﷐﷐﷐3﷮2﷯﷐𝑥﷮2﷯ − ﷐1﷮3𝑥﷯﷯﷮6﷯ Putting n = 6 , a = ﷐3﷮2﷯﷐𝑥﷮2﷯ , b = −﷐1﷮3𝑥﷯ Tr+1 = 6Cr ﷐﷐﷐3﷮2﷯﷐𝑥﷮2﷯﷯﷮6 − 𝑟﷯﷐﷐−﷐1﷮3𝑥﷯﷯﷮𝑟﷯ = 6Cr ﷐﷐﷐3﷮2﷯﷯﷮6 − 𝑟﷯ (﷐𝑥﷮2﷯﷐)﷮(6 − 𝑟)﷯ ﷐﷐−﷐1﷮3﷯× ﷐1﷮𝑥﷯﷯﷮𝑟﷯ = 6Cr ﷐﷐﷐3﷮2﷯﷯﷮6 − 𝑟﷯ (𝑥﷐)﷮2(6 − 𝑟)﷯ ﷐﷐−﷐1﷮3﷯﷯﷮𝑟﷯﷐﷐﷐1﷮𝑥﷯﷯﷮𝑟﷯ = 6Cr ﷐﷐﷐3﷮2﷯﷯﷮6 − 𝑟﷯ (𝑥﷐)﷮12 − 2𝑟﷯ ﷐﷐−﷐1﷮3﷯﷯﷮𝑟﷯﷐﷐𝑥﷯﷮−𝑟﷯ = 6Cr ﷐﷐﷐3﷮2﷯﷯﷮6 − 𝑟﷯ ﷐﷐−﷐1﷮3﷯﷯﷮𝑟﷯ (𝑥﷐)﷮12 − 2𝑟﷯ ﷐﷐𝑥﷯﷮−𝑟﷯ = 6Cr ﷐﷐﷐3﷮2﷯﷯﷮6 − 𝑟﷯ ﷐﷐−﷐1﷮3﷯﷯﷮𝑟﷯ (𝑥﷐)﷮12 − 2𝑟 − 𝑟﷯ = 6Cr ﷐﷐﷐3﷮2﷯﷯﷮6 − 𝑟﷯ ﷐﷐−﷐1﷮3﷯﷯﷮𝑟﷯ (𝑥﷐)﷮12 −3𝑟﷯ Step 2 We need to find the term independent of x So, power of x is 0 ﷐𝑥﷮12 − 3𝑟﷯ = x0 Comparing powers 12 – 3r = 0 12 = 3r ﷐12﷮3﷯ = r 4 = r r = 4 Putting r = 4 in (1) T4+1 = 6C4 ﷐﷐﷐3﷮2﷯﷯﷮6 − 4﷯ ﷐﷐−﷐1﷮3﷯﷯﷮(4)﷯ (𝑥﷐)﷮12 −3(4)﷯ T5 = 6C4 ﷐﷐﷐3﷮2﷯﷯﷮2﷯ ﷐﷐1﷮﷐3﷮4﷯﷯﷯ (𝑥﷐)﷮12 −12﷯ = 6C4 ﷐﷐﷐3﷮2﷯﷮﷐2﷮2﷯﷯﷯﷐﷐1﷮﷐3﷮4﷯﷯﷯ (𝑥﷐)﷮0﷯ = 6C4 ﷐﷐1﷮﷐2﷮2﷯﷯﷯﷐﷐﷐3﷮2﷯﷮﷐3﷮4﷯﷯﷯ (1) = 6C4 ﷐﷐1﷮﷐2﷮2﷯﷯﷯﷐﷐1﷮﷐3﷮4−2﷯﷯﷯ = 6C4 ﷐﷐1﷮﷐2﷮2﷯﷯﷯﷐﷐1﷮﷐3﷮2﷯﷯﷯ = ﷐6!﷮4!﷐6−4﷯!﷯ ﷐﷐1﷮4﷯﷯﷐﷐1﷮9﷯﷯ = ﷐6!﷮4!﷐2﷯!﷯ ﷐﷐1﷮4﷯﷯﷐﷐1﷮9﷯﷯ = ﷐6(5)(4)!﷮4!﷐2﷯!﷯ ﷐﷐1﷮4﷯﷯﷐﷐1﷮9﷯﷯ = ﷐5 ﷮12﷯ Hence, the term which is independent of x is T5 = ﷐5﷮12﷯

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.